Math problem help? please

What is the sum of this geometric series?

(if necessary use the slash bar to enter fractions, reduce fractions to their lowest terms.)

1/4 - 1/8 + 1/16 - 1/32 + ...

Please help!

Thanks:)
~nathan

Each term is (-1/2)^(n + 1) check

Term 1 is (-1/2)*(1 + 1) = (-1/2)^2 = 1/4
Term 2 is (-1/2)*(2 + 1) = (-1.2)^3 = -1/8 and so on

The sum x = 1 to infinity (-1/2)^(n+1)

This is a geometric series without the first two terms which are (1 and - 1/2) = 1/2

The geometric series converges to 1/(1 - -1/2) = 1/3/2 = 2/3 and we subtract the omitted terms 1/2 to obtain 4/6 - 3/6 = 1/6 so the series above converges to 1/6.

1/4 - 1/8 + 1/16 - 1/32... = x
where x is the sum of the series. If you divide each side by 2, you get:
1/8 - 1/16 + 1/32 - 1/64... = x/2
If you subtract the second equation from the first, you get:
x/2 = 1/4
If you solve for x, you get:
x = 1/2

The sum of 1/4 - 1/8 + 1/16 - 1/32... is 1/2.

for those four terms
Sn=a(1-r^n)/1-r

r=-1/2 a=1/4
n=4
s4=1/4(1-(-1/2)^4)/1-(-1/2)
s4=use your calculator

sum to infinity
s=a/1-r
s<>=(1/4)/1-(-1/2)
s<>=use your calculator

first make into common denominator, in this case 32
1/4 turns into 8/32
1/8 turns into 4/32
1/16 turns into 2/32
and 1/32 stays the same

8/32 - 4/32 + 2/32 - 1/32
4/32 + 2/32 - 1/32
6/32 - 1/32
5/32 is your final answer